Question Number 133840 by Algoritm last updated on 24/Feb/21 Answered by Dwaipayan Shikari last updated on 24/Feb/21 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{{x}} \Gamma\left({x}+\mathrm{1}\right)−\mathrm{4}\Gamma\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}=\frac{{x}^{{x}} \left({logx}+\mathrm{1}\right)\Gamma\left({x}+\mathrm{1}\right)+\Gamma'\left({x}+\mathrm{1}\right){x}^{{x}} −\mathrm{12}\Gamma'\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma'\left({x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)+\mathrm{4}\Gamma'\left(\mathrm{3}\right)−\mathrm{12}\Gamma'\left(\mathrm{3}\right)}{\mathrm{3}\Gamma'\left(\mathrm{3}\right)}=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)}{\mathrm{6}\left(\frac{\mathrm{3}}{\mathrm{2}}−\gamma\right)}−\frac{\mathrm{8}}{\mathrm{3}} \\…
Question Number 133830 by metamorfose last updated on 24/Feb/21 $${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−{cos}\left({x}\right){cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right){cos}^{\mathrm{3}} \left(\mathrm{3}{x}\right)…{cos}^{{n}} \left({nx}\right)}{{x}^{\mathrm{2}} }=…? \\ $$ Answered by EDWIN88 last updated on 24/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 2735 by prakash jain last updated on 25/Nov/15 $$\mathrm{Does}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{converge}? \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\mid\frac{\mathrm{sin}\:{i}}{{i}}\mid \\ $$ Commented by Filup last updated on 26/Nov/15 $$\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{limit}…
Question Number 2722 by prakash jain last updated on 25/Nov/15 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{convergent} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{sin}\:{i}}{{i}} \\ $$ Answered by Filup last updated on 25/Nov/15 $$\mathrm{no}…
Question Number 133758 by EDWIN88 last updated on 24/Feb/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}/\mathrm{3}} }\right)\left(\mathrm{tan}^{−\mathrm{1}} \sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\mathrm{tan}^{−\mathrm{1}} \sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:\right)=? \\ $$ Answered by liberty last updated on 24/Feb/21 $$\left(\mathrm{i}\right)\:\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}/\mathrm{3}}…
Question Number 68178 by Mikael last updated on 06/Sep/19 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left({cosx}\right)^{{sin}\mathrm{2}{x}} −\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$ Commented by mathmax by abdo last updated on 06/Sep/19 $${let}\:{f}\left({x}\right)\:=\frac{\left({cosx}\right)^{{sin}\left(\mathrm{2}{x}\right)}…
Question Number 2624 by prakash jain last updated on 23/Nov/15 $$\mathrm{Old}\:\mathrm{question}\:\mathrm{related}\:\mathrm{to}\:\mathrm{greatest}\:\mathrm{int}\:\mathrm{function}. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}+{x}\rfloor=\mathrm{1} \\ $$$$\lfloor\mathrm{1}\rfloor=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor=? \\ $$ Answered by Filup last…
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Question Number 133583 by bemath last updated on 23/Feb/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{24}}}{\mathrm{x}^{\mathrm{6}} } \\ $$ Answered by EDWIN88 last updated on 23/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 67996 by TawaTawa last updated on 03/Sep/19 Commented by mathmax by abdo last updated on 03/Sep/19 $${we}\:{have}\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow{n}^{\mathrm{2}} +\mathrm{1}\leqslant{n}^{\mathrm{2}} \:+{k}\leqslant{n}^{\mathrm{2}} \:+{n}\:\Rightarrow \\ $$$$\sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}\leqslant\sqrt{{n}^{\mathrm{2}}…